The key is your speed around the black hole. You need to go extremely fast to make one hour equivalent to 7 years. I don't know the equations, though.

Some info on the black hole in Interstellar here. For it to work the black hole needs to be extremely large and be spinning extremely fast. Something found in the centre of galaxies.

The key is your speed around the black hole. You need to go extremely fast to make one hour equivalent to 7 years.

Gravity can cause time dilation by itself. Despite their significantly higher speed, clocks on GPS satellites actually tick faster than those on Earth, because the decreased gravitational influence more than offsets the increased velocity.

Merely being close to a black hole would slow time for you, even if you were (somehow) not moving. Of course, in practice, velocity time dilation must also be factored in as you must orbit the black hole.

All forum users, please read this! My SE mods and addons Phenom II X6 1090T 3.2 GHz, 16 GB DDR3 RAM, GTX 970 3584 MB VRAM

Time dilation is not simply a function of your speed (derivative of your position in 3-space with respect to coordinate time). This is only true in flat space-time, which falls into the realm of special relativity. You can also derive the equation for this fairly easily, using the 4D version of Pythagorus' Theorem, and some algebra. (Maybe I'll show this later, because it's a fun example which illustrates foundations of special relativity.)

In curved space-time, figuring out time dilation is a lot harder. Instead of velocity through space, you need velocity through space-time (derivative with respect to proper time, tau), and the metric describing the space-time geometry.

I'll go over how to do this in detail later. (Celebrating my Dad's birthday today). Also, I'm not going to do it properly, because Gargantua (the black hole in Interstellar) rotates. A rotating black hole is described by the Kerr metric, which is significantly more complicated than the Schwarzschild metric (describing an uncharged, non-rotating black hole).

For now, I'll say that time dilation effects from a non-rotating black hole would not be significant enough to match the film's description. You'd have to be extremely close to the event horizon, where there are no stable orbits. But for a rapidly rotating black hole like Gargantua, the spin is very important. It literally drags space-time around the hole, an effect called "frame dragging". Kip Thorne wanted the science of Interstellar to be as accurate as possible, and he actually did use correct equations describing this for the movie. How closely the film's use of time dilaton follows it, I don't know, but it's probably not far off! From what I've seen of discussion by other astrophysicists, you actually can have stable orbits with such large time dilation factors.

That is really cool. A couple descriptive errors (e.g. describing how to make a black hole), but otherwise great info. The graphic of Gargantua really helps show how gargantuan it truly is. That is a monster black hole! For comparison, the black hole at the center of our galaxy fits easily inside the orbit of Mercury.

Also, I think I'll post the derivation for the special relativistic time dilation formula here tonight, the one that assumes you're in flat space-time (or at least flat enough that contribution from gravitational field is negligible for your desired precision). It's a little bit of a grind, but the math itself is surprisingly easy considering we're talking about something most people view as arcane -- the relativity of time and space. But if you know even the bare bones of geometry and algebra, you can do it yourself and impress your friends.

Start with a thought experiment.

Consider the following apparatus. Let’s build a box. Inside is a laser emitter, aimed at a detector on the other side, some known distance away. The laser activates, the beam travels across the box, and hits the detector. The span of time it takes for this to happen makes for one tick of what we will call a clock. When the detector receives the laser pulse, the laser may fire again. Actually, heck, let's idealize this even further -- instead of a laser and detector, have a single photon bouncing between two perfect mirrors. Each bounce corresponds to a tick of the clock.

How long do these ticks take? First let’s look at the frame of reference moving with the clock – the rest frame. Easy: the time it takes for the pulse to cover the distance is simply the distance divided by the speed of the pulse: t=d/c. Or, d=ct_{r}, where r indicates that this is the rest frame we're dealing with.

What if the clock is moving relative to the observer (let’s say perpendicular to the line connecting the mirrors)? The observer follows the path of the photon as it bounces back and forth, and notices that, in the time between bounces, the mirrors have moved a distance vt_{m} (m for moving frame), owing to the motion of the clock itself. For the photon to move from one mirror to another, it must follow an angled path. The faster the clock is moving, the more severe that angle. The distance covered is thus greater as well.

So, in the moving frame, the clock moves a distance vt_{m}, and the distance travelled by the photon is ct_{m}.

The Newtonian thinker might find this absurd, that instead it is the speed of light that must change. But we know from experiment that this is not true! The speed of light is a constant for everyone, regardless of the state of motion. If c is constant, then d and t must vary. So in both frames, we use the same c, and we’ll calculate the change in t.

Here are our observations:

-In the rest frame, the clock moves a distance zero, by definition. -Again in the rest frame, the photon covers a distance ct_{r}. -In the moving frame, the clock moves a distance vt_{m} -Again in the moving frame, the photon covers a distance ct_{m}

We must find a formula relating all of these. As usual in solving problems of a geometric nature, it is helpful to draw a picture:

This is a right triangle! You know what to do:

By the Pythagorean Theorem, equate the square of the hypotenuse to the sum of squares of the other two sides. We obtain

The rest is just algebraic manipulation. Admittedly, this is not so obvious at first, and it takes a few steps. Ultimately, we want to isolate the terms involving time.

and finally, taking the square root and defining t_{m} over t_{r} as the time dilation factor, we find:

Plotting it (and setting c=1), we get the classic asymptotic curve. Special relativistic time dilation is minuscule for every-day speeds, but grows enormously as you approach the speed of light.

One may very well ask if this result is valid, since we started with a pretty contrived definition of a clock which itself uses the speed of light. I assure you, it doesn't matter. Any clock you could possibly conceive of will exhibit this time dilation effect in exactly the same way -- digital clocks, mechanical, atomic, biological (your heart; brain), whatever you want. The effect is demonstrable in numerous experiments, and as HarbingerDawn noted earlier, it is important to the proper functioning of GPS! If we did not account for relativity (special and general), GPS would incur a rapidly accumulating error and be rendered useless.

More to follow on general relativity, including time dilation for a circular orbit around a non-rotating black hole, next time!

If not moving fast enough to be in orbit, I think the main concern would be the experienced G forces, not the experienced time dilation.

I've one issue with Miller's planet: I suppose it must have been formed elsewhere and then found an orbit close to the black hole, otherwise it would be less than 200,000 years old.

How would it be to observe the sky from Miller's planet?

So this planet should orbit just in 80 meters above the event horizon of the black hole! Taking into account dime dilation due to speed, which must be close to speed of light, might increase this distance. But anyway this planet could not exist - it simply will be rip apart by tidal forces.

And before that, the people on it would be instantly dead, and at least the sea would looks like ice with very powerful volcanos due to tidal forces. An increase of "just" 30% of gravity (like in the movie) just doesn't fit with a 7 years-1 hour time dilatation.

And same thing for the other planets, a monster like gargantua with such a gravity would crush anything around it.

So this planet should orbit just in 80 meters above the event horizon of the black hole! Taking into account dime dilation due to speed, which must be close to speed of light, might increase this distance. But anyway this planet could not exist - it simply will be rip apart by tidal forces.

Have you taken into account that the black hole is spinning at 0.998c?

But actually fidal forces depends on the mass of the black hole.

An 8 solar masses black hole would kill you way before you could reach event horizon, but a black hole with 10^8 solar masses could let you cross the event horizon with "no problems". Actually, a planet behaves differently from an human body, but being such a massive object means that tidal forces becomes disruptive at a very low distance to the center (compared to event horizon radius), so... I don't know, maybe the planet could resist and behave like Io does related with Jupiter...

The universe is not required to be in perfect harmony with human ambition.

@SpaceEngineer: The BadAstronomer made the same criticism himself, but then realized that Gargantua is a rotating black hole. This increases the severity of time dilation effects. Apparently -- though I've not seen it shown explicitly -- you can have stable orbits with arbitrarily large gamma in Kerr geometry. Tidal effects are probably still lethal to the planet, though. If not only from the radial derivative of gravitational field, then probably by frame dragging.

@Midtskogen: A black hole "spinning at some fraction of c" makes no physical sense. They should say what the angular momentum is, or what percentage of maximally rotating. We could assume they mean 99.8% of a maximally rotating black hole.

Also, wow, yeah, whoever made that graphic did forget a factor of 2 in there when figuring its size. Or if they wish to mean it has a radius of 1AU, then the mass needs to be half of that -- 50 million solar masses.

I hope that a double post can be forgiven in this case. Also, this post is monstrous, probably the longest post I have ever written anywhere. I hope you can forgive me for that, too.

Earlier I showed a derivation for special relativistic time dilation – the slowing of clocks due to relative motion of observers.

Now I will bring gravity into the picture, and show the time dilation for (a) A body maintaining constant position in a gravitational field. (b) A body in a circular orbit.

For the above, I will assume the gravitational field outside a spherically symmetric, non-rotating distribution of mass, or Schwarzschild geometry. This is a good first approximation for the field of the Earth, a star, or a non-rotating black hole. But I will also discuss (without derivation) the time dilation for a circular orbit of a rotating (Kerr) black hole, and the consequences for the planet in the film Interstellar (re: Salvo’s question from the last page).

For most, these derivations are going to be superfluous. They are purely for any who are curious about where the formulas come from. I tried very hard to make that process as clear as possible, but I must warn it may still be pretty intense, with a fair amount of mathematics and notation introduced along the way. General Relativity is no walk in the park. If it is too much, just skip through it for the final formulas and discussion.

Finally, this is NOT going to be a thorough or rigorous review. It’ll be lengthy though, so I’ll spoilerize it in sections so that it is not too enormous upon loading.

Section 1: Metrics, Tensors, and the Line Element Wherein the reader remembers an old friend – Pythagorus’ Theorem – and, after equipping him with new tools, sends him forth to measure the heavens.

How does one find the distance between two points in space, say on the x-y plane, but without the use of a ruler to reach between them? Simple: start at one point and mark off distance along the x axis, then again along the y axis, until you reach the second point. You’ve drawn the legs of a right triangle, with the distance being the hypotenuse. You know the rest. By Pythagorus’ Theorem, the distance (let’s call it ‘s’) is given by:

Or in differential form:

This expression is an example of what we call a line element. It tells us the squared length of an infinitesimal displacement in terms of coordinate components. In this case, the coordinates x and y. Later, when we look at the geometry of curved space-times, we will want to be able to find and use the line elements that describe those geometries.

To that end, we’ll make use of a metric. Metric tensor, to be precise. A tensor is a geometric object which is preserved under coordinate transformations. They are a huge help in physics, where we want to be able to describe things independently of choice of coordinates or reference frames. A metric tensor can be thought of as a mathematical machine which, when fed a vector, spits out its squared length. (It’s actually performing the dot product). Here’s what it looks like, in the case of a flat, 2D Euclidean space, and being fed a displacement vector ξ "Xi":

Notice the sub and superscripts. These are not exponents, but indices. Here, they range from 1 to 2 (the dimensions of the space). That each index is repeated (one upper, one lower), means that they are to be summed over (Einstein’s summation convention). This convention helps us write what would otherwise be long, tedious expressions in a very compact form.

The components of the metric (g_{11}, g_{12}, etc) determine the properties of the space. Let’s write them out explicitly: g_{11} = g_{22} = 1 g_{12} = g_{21} = 0.

This notation might remind some of the unit matrix, where the diagonal entries are 1s and the rest 0s. Indeed, the components of the metric are often written in matrix form:

Multiplying each entry by the associated coordinate (here x and y), we get the line element described by the metric:

And that of course is the Pythagorean theorem again. We’ve just shown that this theorem arises naturally out of the metric describing flat, Euclidean 2D space.

It may seem insane that we go to these great lengths to complicate such a simple expression. The metric components here are just 1’s and 0’s, so a bunch of the terms vanish from the line element, giving us back ds^2 = dx^2 + dy^2. Brilliant. So why do we invoke such complex mathematical notation for this? The answer is because although the components are very simple here, they are not that simple in general. Space-time is not Euclidean, and it in general is not flat, either, so the metrics in general relativity can be quite complicated. Let’s show this with one more example before moving on to the Schwarzschild metric. Schwarzschild geometry is curved, so let’s look at flat space-time, AKA the Minkowski space of special relativity. Minkowski space is flat, but it is not Euclidean.

In matrix form, the components of Minkowski space are:

Which with coordinates [t,x,y,z] gives us a line element of:

(Aside: I’m using the -+++ “space-like” convention here. It could be the other way (+--- or “time-like”). It doesn’t matter which we use as long as we are consistent).

This is definitely not your typical high-school geometry anymore. Notice we’ve just found a way for a squared length to be negative! If the time interval is larger than the space intervals, then the sum is negative and so ds^2 is negative.

Incidentally, the sign of ds^2 is a wonderful diagnostic for determining different regions of space-time. Again using the -+++ convention, a negative ds^2 is called a time-like curve, which describes velocities less than c, or any allowed observer. ds^2 = 0 is the speed of light, or a null curve, and ds^2 > 0 is a space-like curve, which one must travel faster than light to follow. Thus, events separated by a space-like interval cannot be causally related.

And this ends the brief overview of metrics, tensors, and space-time geometry. Next we’ll look into the metric of a black hole, and use it to calculate the time dilation near one.

Section 2: The Schwarzschild Metric Wherein the reader, now armed with magic potions and incantations of geometrodynamics, confronts the darkness of the black hole.

We’re ready for the Schwarzschild metric. Since this involves spherical symmetry, it is useful to replace the Cartesian coordinates [t,x,y,z] with Spherical coordinates [t,r,θ,φ]. Lowercase t is coordinate time, r is radial distance, θ is polar angle (analogous to latitude), and φ is azimuthal angle (longitude). Here are the Schwarzschild metric components in matrix form:

And as a line element:

Yikes! This looks complicated, indeed. The components are no longer simply numerical 1s, -1s and 0s, but functions, all of which depend on r. Since this metric assumes being outside of a spherically symmetric mass distribution, it follows from Gauss’ theorem that we can treat this distribution as being a singularity – all the mass contained at a point at the center. Also, note that I’m using “geometrized” units, where G=c=1. We do this a lot in relativity to make formulas more manageable. For example, here “2M” is actually the Schwarzschild radius, 2GM/c^{2}. Geometrized units can be weird. For example, you can say that the mass of the Sun is 1.79x10^{47}joules (multiply by c^{2}), or 1.48 kilometers (multiply by G/c^{2}).

One more aside. Notice that as r approaches 2M, the metric goes to infinity, and is undefined at r=2M. For a while after the Schwarzschild metric was derived, physicists thought this actually represented a singularity – that space-time became singular at the black hole’s horizon. But this is not a true singularity, but a coordinate one. The geographic poles, in lat/lon coordinates, are also coordinate singularities. There is nothing weird about the geometry there, and the singularity can be removed by transforming into a new coordinate system. An object falling into a black hole does not stop or get crushed out of existence at the horizon, but instead sails right through it in a finite amount of time (as seen from the object itself). However, there is a true space-time singularity at the black hole’s center, where objects do get crushed into oblivion.

Okay so it is not pretty, but we do have a metric. What do we do with it? What we want is to find the time dilation, first for a clock maintaining a fixed position, and then for a clock in a circular orbit. The comparison clock is one that is very far away, where the space-time asymptotically approaches the flatness of Minkowski.

To achieve this, we need to understand the motion of the clocks within the gravitational field. We do this with the 4-velocity. Where 3-velocity is the change of position (in 3-space) with respect to time, 4-velocity is the change of position (in space-time) with respect to proper time, tau. What is proper time? That’s what you what you read off (with appropriate signs used) from the line element, ds^2. All clocks tick off proper time. This may make some sense after reflecting on the space-like, time-like, and null curves discussed earlier. If ds^2 = 0 is a light-like curve, then that means a clock travelling at the speed of light ticks zero times. Time does not pass for something moving at c.

The 4-velocity is defined (here in Schwarzschild spherical coordinates) as

By definition, the ratio of coordinate time (t) to proper time (tau) equals gamma – the time dilation factor. Gamma is the quantity we want to find at the end of the day.

One (and probably the easiest) way to do that is to make use of the metric tensor, inputting the 4-velocity vector to produce an invariant quantity. The squared length of a 4-velocity in relativity is always -1. (Again, in geometrized units.)

Writing this out explicitly in coordinate form (recall the line element / metric components of Schwarzschild given above), we have:

This really does look like a nightmare. But panic not. Remember we are looking at a clock which is not moving through the space. That means r, theta, and phi are all constant, so their derivatives are zero and a bunch of these terms will vanish:

Niiiiice. This is much simpler and we can solve for gamma immediately:

Sure enough, this is the formula SpaceEngineer used a couple posts ago. Also, remember that 2M is the geometrized Schwarzschild radius, 2GM/c^{2}. This expression is true for any spherically symmetric mass, too. Not just black holes. If you calculate this radius for the Earth (it’s a little under 1cm) and plug in the true radius (6371km) for r, you’ll get the time dilation due to gravity at Earth’s surface, with respect to someone very far away. Naturally, it’s very small, but it has been measured successfully using extremely precise atomic clocks at different altitudes.

For a black hole, the Schwarzschild radius coincides with the event horizon (if it isn’t spinning). Here, the time dilation with respect to a distant observer goes infinite. If you fall into a black hole, someone watching you from afar will never see you reach the horizon for this reason. You will seem to slow down (and rapidly fade from view due to gravitational redshift [which is really the same effect as the time dilation]) as you get closer. In practice, this fadeout happens extremely quickly, within a fraction of a second.

From your perspective, however, you pass through the horizon without noticing anything weird at all. Time does not seem to pass slowly to you, because every part of you is slowed in the exact same way. The proper time of your path to the singularity is finite. You will still never see the singularity, though, because no light rays can come off of it. Even from a millimeter away, it is invisible to you. You also won’t be alive when you meet the singularity, since tidal forces will destroy you first. For a small black hole, you won’t even reach the horizon intact. The lethal radius of tidal forces grows as the cube root of mass, but the event horizon radius grows faster -- linear with mass. For a supermassive black hole weighing millions of suns, the event horizon is much larger than the lethal radius.

A planet does not fare as well as a human would, because a planet is much larger. Therefore the difference in gravitational pull on one side vs. the other is also larger. A planet would be torn apart easily even when a human does not yet feel significant tidal force. Similarly, a bacterium would fare better than a human. That said, everything, no matter how small, gets torn apart somewhere before the singularity. At a singularity, tidal forces are infinite, at least according to classical theory (future ‘quantum gravity’ may fix this).

While on the subject of what it’s like to be near/inside a black hole, let me dispel a very common misconception. A lot of people think that as you fall through the event horizon, you become enveloped in darkness. This is shown that way in Interstellar, as well as in a documentary series on black holes from a few years back. It is wrong. If you could somehow slowly lower your way into a black hole, then it is true that your view of the outside universe is compressed to a smaller and smaller area above you. But that is only because you are accelerating like mad to remain in place against the pull of gravity. That causes relativistic beaming, concentrating the view in the direction you accelerate (upward, against the inflowing rush of space – think of a black hole as like a waterfall of space).

But if you freely fall into a black hole, then you do not see this effect. The black disk that you see is also not the true “event horizon”, but the “antihorizon”. The event horizon is invisible and not noteworthy to you. There is no obvious way to tell that you’ve passed through it. The black antihorizon however always seems to be some distance ahead of you. It grows as you approach, but only takes up 180 degrees of your view when you finally hit the singularity.

So falling into a black hole doesn’t look like falling through a black surface and into an abyss – it looks instead more like approaching the surface of a black sphere and landing (and getting crushed to death) on it. The view of the universe outside the hole is obviously extremely distorted, though.

For realistic visualizations of these effects and other black hole and space-time oddities, I highly recommend Andrew Hamilton’s webpage.

Next section, we’ll investigate clocks that are not at rest, but moving around a black hole, and see what that does to the time dilation.

Section 3: Time Dilation of Circular Orbits Wherein the reader, triumphant, lassos the black hole.

Now that we have the Schwarzschild metric and know how to use it, we’re ready to examine time dilation for a circular orbit. Let’s look at the result from putting the 4-velocity into the metric again:

For a stationary clock, we had r, theta, phi constant. For a circular orbit, r is constant, and by spherical symmetry, we can choose the orbit to lie in the equatorial plane (theta=pi/2=constant). Choosing theta=pi/2 is nice because it also makes Sin^{2}(theta) = 1.

This then simplifies to:

Let's pause to look at this for a moment. This expression still contains exactly what we had before from the time dilation of a static clock, but now with an additional term involving d(phi)/d(tau). We interpret this to mean that for a circular orbit, the dilation factor is a combination of the gravitational time dilation, plus a factor from the orbital velocity. In other words, the method we are using here automatically accounts for both special and general relativistic time dilation effects. They both derive from the same underlying principle: motion not just through space, but space-time.

Unfortunately we can't yet solve this for gamma because we don't know what d(phi)/d(tau) is. To replace it, we’ll have to call upon some celestial mechanics. Much of this I will not derive, but will just show the substitutions:

This is basically the Chain Rule of calculus, rewriting d(phi)/d(tau) in terms of dt. d(phi)/dt is Ω, the angular velocity relative to the distant observer, and dt/d(tau) again is gamma, the time dilation factor that we seek.

From mechanics,

where l is the angular momentum and e is the energy. Both of these are conserved quantities, and for a circular orbit, this ratio is

Geometrizing these expressions and substituting back up:

Finally we have the expression for d(phi)/d(tau), and can plug back in to our equation and solve for gamma:

And at last,

A lot of work for a simple result. And the only thing that changed from the formula for the static clock in Schwarzschild is that the factor of 2 is now a 3. Isn't it amazing how things in math/physics turn out sometimes?

Let’s plot these two equations together, and see how they compare (for a given mass M).

In the case of the static clock, the time dilation approaches infinity at r=2M, or the event horizon. This is of course a completely implausible situation, since the acceleration required to hover close to the horizon is enormous.

But for an orbiting clock, this factor approaches infinity as r approaches 3M, or 1.5 Schwarzschild radii. There is a name for this radius – the photon sphere. There, a photon travelling perfectly tangential to the hole could, hypothetically, remain in circular orbit! (Hypothetically, because such photon orbits are enormously unstable – the smallest perturbation or imperfection of the orbit will cause it to rapidly spiral either into the hole or off to infinity).

A massive particle cannot be in an orbit at this radius, so this infinite time dilation result only applies to photons (which shouldn't come as a surprise, since we know a proper time of zero corresponds to null or light-like curves). It also turns out that this circular orbit result is unphysical for any material body closer than r=6M, or the Innermost Stable Circular Orbit (ISCO). You can have unstable orbits that come within ISCO (as close as the photon sphere, and within that all paths will spiral into the hole), but they are not circular, thus invalidating both the assumption of r=constant and the orbital mechanics substitutions. Orbits within ISCO can look very strange, with a "zoom-whirl" behavior -- zooming around the hole several times near periapsis before whirling back out to apoapsis. A fine example of how Keplerian dynamics break down in strongly curved space-time.

At ISCO, the time dilation factor is the square root of 2, meaning a clock there ticks about 71% the rate as a clock far away. This is impressive but not overwhelmingly so. It amounts to an additional ~25 minutes per hour. Clearly, we cannot get such high time dilation factors as 7 years per hour from orbiting a simple Schwarzschild black hole.

This all changes when we allow for a rotating black hole, which we'll examine next.

Section 4: Further discussion, Kerr geometry, and Miller’s Planet In which the reader encounters whirlpools of space-time.

Guess what, no more enormous derivations! (I know, I’m disappointed, too.)

The Kerr metric, describing geometry around a rotating black hole, is incredibly complicated, and is just way too much to attempt to go over here. Consider that the Schwarzschild solution was found in the same year as Einstein’s publication of General Relativity (1916), whereas Kerr’s solution wasn’t found until 1963. There’s great info on Kerr geometry and rotating black holes on Wikipedia, and those wanting more mathematical depth may enjoy these lecture notes.

The basic idea of Kerr geometry is that coordinate frames are literally dragged around in the direction of rotation. Since the Earth rotates [citation needed], there is frame dragging here, too. It is unbelievably weak, but nevertheless experimentally verified by the Gravity Probe B in 2005, using some of the most finely crafted materials ever made by mankind.

For a spinning black hole, this frame dragging effect is severe. If you, from far away, dropped a stone into one, you would see the stone start to drift to the side, then suddenly spiral around the hole as if trying to attain orbit. (It doesn’t, though, because this zero-angular-momentum trajectory still plunges through the event horizon).

Close to the event horizon, the frame dragging is even faster than light. The boundary where it equals the speed of light is the “static limit”, and the region inside is the ergoregion (some call it ergosphere, but I dislike the name because it is not spherical). The name is derived from the fact that within this region, enormous amounts of energy can be harvested from the black hole’s rotation. Conservation of energy obviously still applies, and the cost is that the black hole loses some of its spin. The faster-than-light behavior of space within the ergoregion also means that it is impossible to remain at rest there, even though you are outside the event horizon. You are forced to move in the direction of rotation. In principle, it is possible to move into the ergoregion, and then escape back out.

There is a maximum amount of spin that a black hole can have, too. A black hole with the maximum spin is said to be extremal or maximally rotating. As the angular momentum of the hole increases, the radius of the event horizon decreases, to a minimum value of half the Schwarzschild radius. Incidentally, this means that SpaceEngineer and myself were wrong about the graphic of Gargantua earlier. Gargantua is very nearly maximally rotating, and so its event horizon radius is indeed about 1AU as the graphic shows. The ergoregion of Gargantua is much larger, reaching ~2AU along its equator.

Time Dilation on Miller’s Planet:

Time dilation at the ISCO for a rotating black hole is greater than for a Schwarzschild black hole. One may think of this as a consequence of the ISCO moving inward as the spin increases. The time dilation factor even approaches infinity as the black hole’s angular momentum goes extremal. Apparently, (I’m taking Kip Thorne’s word for it), a dilation factor of ~60,000 (dilating 1 hour to 7 years) requires the black hole to rotate at about one part in 100 trillion less than its maximum value.

But for that to work, Miller’s planet must be literally right on the edge of the horizon again. It is not depicted this way in the film, probably because Miller’s planet is the first to be explored, and they did not want to show the cool visual effects of being that close to the hole so early in the film. Better to save them for later.

Another problem with this is that black holes probably cannot get that close to maximally rotating in nature. Kip himself showed that a counteracting torque from radiation will offset the increased rotation from accretion, reaching an equilibrium at about 99.8% of the maximum. It seems that for the sake of showing extreme time dilation in the film, they decided to ignore this limitation. (This is also probably where the graphic came up with “rotating at 99.8% of c”, but this description makes no sense. Unlike a rotating rigid body, there is no clear/unique way to define black hole rotation through rotational velocity, and all rotating black holes have regions rotating faster than c.)

So, the time dilation for Miller's planet does make sense, but as mentioned earlier, it's probably not at all realistic to have a planet actually be there. Even if it wasn't tidally disrupted, the amount of radiation given off by the accretion disk (which it must be orbiting within!) would vaporize it. And the Endurance and her crew. Interstellar is deeply rooted in good science, but still takes quite a few liberties for aesthetic and plot.

And that, dear folks, is it. If you stuck through all of that, good for you! I did spend a lot of time writing, revising, and contemplating how to make this post, but it was an enjoyable exercise. And I hope it was interesting for you as well.

Cheers.

Added: I just realized a couple of the formulas have both capital R and lowercase r. That is a typo -- they should be the same (lowercase) r, representing radius from the singularity.

Watsisname, I want to know you IRL so we can be best friends. With the effort you put in these posts I suspect that discussions with you would be very interesting. These last few posts are great.

Intel Core i7-5820K 4.2GHz 6-Core Processor G.Skill Ripjaws V Series 32GB (4 x 8GB) DDR4-2400 Memory EVGA GTX 980 Ti SC 6GB

By the way, I saw Kip's The Science of Interstellar on the shelf at the bookstore today. I was skeptical of it at first, thinking it would probably be pretty hokey, but picked it up and started skimming to see. It is actually extremely good! Thorough, excellently written, beautifully illustrated, and easily understandable to popular audiences. Lots of cool science involved that I hadn't even realized (he really did do his homework), plus interesting discussion of his experience working with the rest of the team in Hollywood. Recommended!